보존 벡터장

$4\,\times \,{10}^{29}\text{erg}$

${\int }_{C}\text{F}\cdot d\text{r}\approx 2.9923$

$\text{circulation}=\pi {a}^{2}\,\text{and flux}=0$

Let $\varphi$ be a potential for $\mathbf{F}$. Define $\phi=\varphi+ax+by+cz$. Then ${ \boldsymbol{\nabla} } \phi = { \boldsymbol{\nabla} }\varphi+(a,b,c)=\mathbf{F}+(a,b,c)$. So, $\mathbf{F}+(a,b,c)$ is also conservative.

a. If $\mathbf{F}+\mathbf{G}$ is conservative for any particular $\mathbf{F}$ and $\mathbf{G}$, then by definition, there exists a potential $\varphi$ with $\mathbf{F}+\mathbf{G} = { \boldsymbol{\nabla} } \varphi$.

Since $\mathbf{F}$ is conservative, there also exists a potential $\psi$ with $\mathbf{F} = { \boldsymbol{\nabla} } \psi$.

But now $\mathbf{G} = (\mathbf{F}+\mathbf{G})-\mathbf{F}={ \boldsymbol{\nabla} } \varphi - { \boldsymbol{\nabla} } \psi = { \boldsymbol{\nabla} }(\varphi-\psi)$. That means the function $(\varphi-\psi)$ is a potential for $G$. However, this is impossible: since $\mathbf{G}$ is non-conservative, no function with this property exists.

So it is not possible that $\mathbf{F}+\mathbf{G}$ is conservative. It must be non-conservative. b. Counterexample: if $\mathbf{F} = -\mathbf{G}$, then $\mathbf{F}+\mathbf{G} = \mathbf 0 = { \boldsymbol{\nabla} } c$ for any constant $c$. c. Since both fields are conservative, they both have potentials, say $\mathbf{F}={ \boldsymbol{\nabla} } \varphi$ and $\mathbf{G} = { \boldsymbol{\nabla} } \psi$. Then $\mathbf{F}+\mathbf{G} = { \boldsymbol{\nabla} }\varphi+{ \boldsymbol{\nabla} }\psi={ \boldsymbol{\nabla} }(\varphi+\psi)$. That is, $(\varphi+\psi)$ is a potential for $\mathbf{F}+\mathbf{G}$, so $\mathbf{F}+\mathbf{G}$ is conservative.

If $\varphi$ is a potential for $\mathbf{F}$, then:

• $\frac{\partial \varphi}{\partial x}=x+y$, so $\varphi = \frac{1}{2}x^2+xy+\psi_1(y)$
• $\frac{\partial \varphi}{\partial y}=x-y$, so $\varphi = xy-\frac{1}{2}y^2+\psi_2(x)$

So, for instance, $\varphi = \frac{1}{2}x^2+xy-\frac{1}{2}y^2$ is a potential for $\mathbf{F}$.

If $\varphi$ is a potential for $\mathbf{F}$, then:

• $\frac{\partial \varphi}{\partial x}=\frac1x-\frac1y$, so $\varphi = \ln|x|-\frac{x}{y}+\psi_1(y)$
• $\frac{\partial \varphi}{\partial y}=\frac{x}{y^2}$, so $\varphi = -\frac{x}{y}+\psi_2(x)$

So, for instance, $\varphi=\ln |x| - \frac{x}{y}$ is a potential for $\mathbf{F}$.

If $\varphi(x,y,z)$ is a potential for $\mathbf{F}(x,y,z)$, then:

• $\frac{\partial \varphi}{\partial x}(x,y,z)=\frac{x}{x^2+y^2+z^2}$, so $\varphi(x,y,z) = \frac12\ln(x^2+y^2+z^2)+\psi_1(y,z)$
• $\frac{\partial \varphi}{\partial y}(x,y,z)=\frac{y}{x^2+y^2+z^2}$, so $\varphi(x,y,z) = \frac12\ln(x^2+y^2+z^2)+\psi_2(x,z)$
• $\frac{\partial \varphi}{\partial z}(x,y,z)=\frac{z}{x^2+y^2+z^2}$, so $\varphi(x,y,z) = \frac12\ln(x^2+y^2+z^2)+\psi_2(x,y)$

So, for instance, $\varphi(x,y,z)= \frac{1}{2}\ln(x^2+y^2+z^2)$ is a potential for $\mathbf{F}(x,y,z)$.

(a) We shall show that $\mathbf{F}(x,y,z)$ is conservative by finding a potential for it. $\varphi(x,y,z)$ is a potential for this $\mathbf{F}$ if and only if

$$\begin{aligned} \frac{\partial \varphi}{\partial x}(x,y,z) &= x \\ \frac{\partial \varphi}{\partial y}(x,y,z) &= -2y \\ \frac{\partial \varphi}{\partial z}(x,y,z) &= 3z \end{aligned}$$

Integrating the first of these equations gives

$$\varphi(x,y,z) = \frac{x^2}{2} + f(y,z)$$

Substituting this into the second equation gives

$$\frac{\partial f}{\partial y}(y,z) = -2y$$

which integrates to

$$f(y,z) = -y^2+ g(z)$$

Finally, substituting $\varphi(x,y,z) = \frac{x^2}{2} -y^2 + g(z)$ into the last equation gives

$$g'(z) = 3z$$

which integrates to

$$g(z) = \frac{3}{2} z^2 +C$$

with $C$ being an arbitrary constant. So, $\mathbf{F}(x,y,z)$ is conservative and $\varphi(x,y,z)=\tfrac{1}{2} x^2-y^2+\frac{3}{2}z^2$ is one allowed potential.

(b) The field $\mathbf{F}= F_1\,\hat{\boldsymbol{\imath}} + F_2\,\hat{\boldsymbol{\jmath}}$ can be conservative only if it passes the screening test

$$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$$

In this case

$$\frac{\partial F_1}{\partial y} =\frac{\partial}{\partial y}\Big(\frac{x}{x^2+y^2}\Big) =-\frac{2xy}{{(x^2+y^2)}^2}$$

is different from

$$\frac{\partial F_2}{\partial x} =\frac{\partial}{\partial x}\Big(\frac{-y}{x^2+y^2}\Big) =\frac{2xy}{{(x^2+y^2)}^2}$$

for all $(x,y)$ with $x$ and $y$ both nonzero. So $\mathbf{F}$ is not conservative.